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02 Jul 2010, 04:38

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Is \(x^2y^3z\)>0?1. \(yz>0\)2. \(xz<0\)

a. Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficientb. Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficientc. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficientd. EACH statement ALONE is sufficiente. Statements (1) and (2) TOGETHER are NOT sufficient

Statement (1) by itself is insufficient. It allows \(x=0\) .Statement (2) by itself is insufficient. It allows \(y>0\) and \(y<0\) .Statements (1) and (2) combined are sufficient. S1 and S2 give that is not 0 and thus \(x^3y^2z\)>0 .The correct answer is C.

Can someone please simplify this for me? I am completely stumped even after reading the explanation

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02 Jul 2010, 05:55

yeahwill wrote:

Is \(x^3y^2z\)>0?1. \(yz>0\)2. \(xz<0\)

a. Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficientb. Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficientc. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficientd. EACH statement ALONE is sufficiente. Statements (1) and (2) TOGETHER are NOT sufficient

Statement (1) by itself is insufficient. It allows \(x=0\) .Statement (2) by itself is insufficient. It allows \(y>0\) and \(y<0\) .Statements (1) and (2) combined are sufficient. S1 and S2 give that is not 0 and thus \(x^3y^2z\)>0 .The correct answer is C.

Can someone please simplify this for me? I am completely stumped even after reading the explanation

Because x has an odd exponent, x^3 can be negative or positive or 0Because y has an even exponent y^2 will be positive or 0z can be positive or negative or 0

1. yz > 0 means both are positive or both negative but we know nothing about x, which can be positive, negative, or 0

2. xz< 0 means one is negative and one is positive which would make you think you have sufficient information because y^2 will also be positive, but it can also be zero in which case you don't have enough information

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Statement (1) by itself is insufficient. It allows \(x=0\) .Statement (2) by itself is insufficient. It allows \(y>0\) and \(y<0\) .Statements (1) and (2) combined are sufficient. S1 and S2 give that is not 0 and thus \(x^3y^2z\)>0 .The correct answer is C.

Can someone please simplify this for me? I am completely stumped even after reading the explanation

You copied question m21q30 incorrectly. Original question m21q30 is below:

Is \(x^2*y^3*z>0\)?

Inequality \(x^2*y^3*z>0\) to be true: 1. \(y\) and \(z\) must be either both positive or both negative, so they must have the same sign (in this case \(y^3*z\) will be positive) AND 2. \(x\) must not be zero (in this case \(x^2\) will be positive).

(1) \(yz>0\) --> \(y\) and \(z\) are either both positive or both negative (first condition). But we don't know about \(x\) (second condition). Not sufficient.

(1) \(yz>0\). If all three are positive than obviously \(x^3y^2z>0\), but if \(y\) and \(z\) are positive and \(x\) is negative then \(x^3y^2z<0\). Two different answers. Not sufficient.

(2) \(xz<0\) --> \(x\) and \(z\) have opposite signs --> \(x^3z<0\). Now \(y^2\) can be either positive or zero so \(x^3y^2z\) can never be positive, it can be either negative if \(y>0\) or zero if \(y=0\). Hence the answer to the question "is \(x^3y^2z>0\)" is NO. Sufficient.

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13 Jun 2012, 11:09

Hi,

\(x^3y^2z>0?\)

Using (1),yz>0,expression reduces to \(x^3y(yz)\), (removing the positive values)or \(x^3y\), we don't know the sign of y. Insufficient.

Using (2) ,xz<0expression reduces to \(x^2(xz)y^2\), (removing the positive values)or xz which is less than zero. But still y can be 0.even in that case the given expression will be less than or equal to 0. Sufficient.

Thus, Answer is (B),

_____________________________________________________\(x^2y^3z>0?\)

Using (1),yz>0,expression reduces to \(x^2y^2(yz)\), (removing the positive values)or \(yz\), which is greater than 0, but x can be equal to zero. Insufficient.

Using (2) ,xz<0expression reduces to \(x(xz)y^3\),or \(xy^3*(-ve)\), no idea about signs of x & y. Insufficient.